1. Giải bất phương trình:  41x>2x
2. Với giá trị nào của m thì phương trình:
      log2+3(x22(m+1)x)+log23(2x+m2)=0 có nghiệm duy nhất
1.41x>2x41x>2x0{2+x>1x2<x1
Đáp số: 5+132<x1
2. Đặt a=2+3 thì PT loga(x22(m+1)x)=loga(2x+m2)
x22(m+1)x=2x+m2>0{x22(m+2)xm+2=0(1)x>2m2
- Nếu Δ=(m+2)2(m+2)=m2+5m+2<0 thì phương trình vô nghiệm
- Nếu Δ=0m=5±172 thì (1) có nghiệm kép x=m+2
PT có nghiệm duy nhất m+2>2m2m>23
Chỉ có giá trị m=5+172 thích hợp
- Nếu Δ>0 thì (1) có hai nghiệm phân biệt x1<x2.PT đã cho sẽ có nghiệm duy nhất {x1<2m2>x2x1=2m2<x2
[f(2m2)<0{f(2m2)=02m2<x1+x22
(f(x)=x22(m+2)xm+2)25m2
Kết luận : Phương trình có nghiệm duy nhất
m=5+172,25m2

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